A swimming pool, 20.0 m ✕ 12.5 m, is filled with water to a depth of 3.73 m. If the initial temperature of the water is 17.5°C, how much heat must be added to the water to raise its temperature to 29.7°C? Assume that the density of water is 1.000 g/mL.

Question

Chemistry

Anabelle Robles

2 December, 02:11

A swimming pool, 20.0 m ✕ 12.5 m, is filled with water to a depth of 3.73 m. If the initial temperature of the water is 17.5°C, how much heat must be added to the water to raise its temperature to 29.7°C? Assume that the density of water is 1.000 g/mL.

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Answers (1)

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Ruby Moss · Answer 1

4.755 x 10¹⁰ J.

Explanation:

The amount of heat absorbed by water = Q = m. c.ΔT.

where, m is the mass of water (m = d x V).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 29.7°C - 17.5°C = 12.2°C).

To calculate the mass of water in the swimming pool, we need to calculate its volume:

Volume of the swimming pool = 20.0 m x 12.5 m x 3.73 m = 932.5 m³ x (10⁶ mL/1.0 m³) = 9.235 x 10⁸ mL.

∴ The mass of water in the swimming pool = d x V = (1.000 g/mL) (9.235 x 10⁸ mL) = 9.235 x 10⁸ g.

∴ The amount of heat absorbed by water = Q = m. c.ΔT = (9.235 x 10⁸ g) (4.18 J/g°C) (12.2°C) = 4.755 x 10¹⁰ J.